Question

How do you graph `f(x)=(1-x^2)/x` using holes, vertical and horizontal asymptotes, x and y intercepts?

Answer 1

no holes
vertical asymptote at `x = 0`
`x`-intercepts at `(-1, 0), (1, 0)`
`y`-intercept is undefined
no horizontal asymptote
slant/oblique asymptote at `y = -x`

Explanation:

Given: `f(x) = (1-x^2)/x`

This is a rational function: `(N(x))/(D(x)) = (a_nx^n+...)/(b_mx^m+...)`

Factor the numerator:
`f(x) = -(x^2 -1)/x = -((x-1)(x+1))/x`

Holes: Since there are no terms that cancel, there are no holes.

Find vertical asymptotes: when `D(x) = 0 => x = 0`

Find x-intercepts by setting `f(x) = 0`:

`0 = (1-x^2)/x`

`0 * x = (1 - x^2)`

`1 - x^2 = 0`

`x^2 = 1`

`x = +- 1`

`x`-intercepts at `(-1, 0), (1, 0)`

Find y-intercept by setting `x = 0 => y =`undefined

Find horizontal asymptote by comparing `m & n`:

If `n < m`, horizontal asymptote is `y = 0`

If `n = m`, horizontal asymptote is `y = a_n/b_m`

If `n > m` there is no horizontal aymptote

In the given equation, `n = 2, m = 1 => n > m` which means there is no horizontal asymptote.

If `n = m+1` there is a slant /oblique asymptote.

You need to use long division to find the slant/oblique asymptote :

`" "-x`
`x |bar(-x^2 + 0x + 1)`
`" "ul(-x^2)`
`" "0x + 1`

`" "color(red)(-x) + 1/x`
`x |bar(-x^2 + 0x + 1)`
`" "ul(-x^2)`
`" "0x + 1` This is the remainder

slant/oblique asymptote is at `color(red)(y = -x)`